how can I simplify this to a phrase without a sigma?!
$$ {1\over r+1}{2r \choose r} + \sum_{i=1}^r \left( {i+1\over r+1} {2r-i \choose r-i}{s+i-2 \choose i} \right) $$ thanks!
2026-03-28 22:25:21.1774736721
Simplify Summation of combination
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There isn’t much you can do. Mathematica tells me the summation equals:$$\frac{2 \binom{2 r-1}{r-1} \left(\, _3F_2 \left[ {2,-r,s-1 \atop 1,-2 r} ;1 \right] -1\right)+\binom{2 r}{r}}{r+1}$$
Where $_3F_2$ is a generalized hypergeometric function.
That basically means: as far as $\sum$s go, this one fits a sort of “nice” form, but there probably won’t be a satisfying closed form.
Here is a table of your entire equation for varying values of $r$ (row number) and $s$ (column number):
$$ \begin{array}{cccccccccc} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ 2 & 5 & 9 & 14 & 20 & 27 & 35 & 44 & 54 & 65 \\ 5 & 14 & 28 & 48 & 75 & 110 & 154 & 208 & 273 & 350 \\ 14 & 42 & 90 & 165 & 275 & 429 & 637 & 910 & 1260 & 1700 \\ 42 & 132 & 297 & 572 & 1001 & 1638 & 2548 & 3808 & 5508 & 7752 \\ 132 & 429 & 1001 & 2002 & 3640 & 6188 & 9996 & 15504 & 23256 & 33915 \\ \end{array} $$
EDIT: It appears your entire equation also equals
$$\frac{s}{r!} \prod_{k=r+1}^{2r-1} (s+k)$$
which might be what you started from in the first place!
EDIT: and of course, assuming $r, s$ are positive, that equals
$$\frac{s}{r} \binom{s+2r-1}{r-1}.$$
I'm seeing if I can find a good proof.